# All simple modules are projective $\Rightarrow$ semisimple [duplicate]

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Let $A$ be a finite dimensional algebra over a field $K$. It is clear that if $A$ is semisimple, then every simple module is projective. Does the converse hold ?

It seems false, but I can't find a counterexample. A non-semisimple algebra with this property must have a non-projective indecomposable module, but that's as far as I could go.

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 23 '15 at 10:03

• It seems to me that every finitely generated $A$-module $M$ has a simple epimorphic image, $S$ say, and that since $S$ is projective, $S$ is isomorphic to a direct summand of $M$. Hence $M$ is isomorphic to direct sum of simple $A$-modules. – Geoff Robinson Jun 21 '15 at 21:15
• The way I did it at the dupe is to observe there are no essential maximal right ideals, and this means all right ideals are direct summands. – rschwieb Jun 23 '15 at 10:09
• I misread your answer there and I'm sorry. But I'm more sorry about your sarcastic tone in the comment above. I'll remove this comment in a few minutes, if you do the same. – egreg Jun 30 '15 at 15:48
• Dear @egreg : No sarcasm was intended at all: I was just illustrating my disbelief with an example, because I did not understand the objection. I frequently use this overly formal format to avoid misunderstandings. In the future please don't assume my intentions are negative. Thanks and Regards – rschwieb Jun 30 '15 at 16:49

Every finitely generated $A$-module has finite length, and therefore it is projective.

Now use a result of Auslander which says the following: the left global dimension of $A$ equals the supremum of projective dimensions of cyclic $A$-modules. This implies that the global dimension of $A$ is zero, hence every $A$-module is projective, so $A$ is semisimple.

$A$ is a module of finite length over itself. Prove, by induction on the length, that a finitely generated module over $A$ is semisimple. In particular $A$ is semisimple.

Note that, if every projective module is simple. Then, in particular, $_AA$ is simple. Thus, $A$ is simple ring. (I am assuming that $A$ is finite dimensional $K$-algebra with identity)

However, if every indecomposable projective left $A$-module is simple, then by using Krull-Schmidt theorem($_AA$ is a direct sum of indecomposable projective modules and these indecomposable summands are uniquely determined up to isomorphism and permutation), $A$ is a direct sum of indecomposable projective left $A$- modules. Hence, direct sum of simple modules. Thus, $A$ is semisimple.

Following the discussion in the comments and considering the title of your question. If every simple module is projective, then by Krull Schmidt theorem and the fact that indecomposable projectives have simple head. We obtain that the simples are isomorphic to their projective cover. Thus, $A$ is a direct sum of simples.

• Who said the projective modules are simple? – user26857 Jun 22 '15 at 15:29
• @user26857 "It is clear that if A is semisimple, then every simple module is projective. Does the converse hold?", what I understood is if every projective is simple, is $A$ semisimple? am I wrong? – Math137 Jun 22 '15 at 15:32
• Please read the title of the question. – user26857 Jun 22 '15 at 15:33
• @user26857 I agree the title is different, but what is written in the question is different from that, isn't it? – Math137 Jun 22 '15 at 15:35